Answer
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Hint: Planck's constant appears in the equation of energy of a photon and the equation of De Broglie wavelength. The dimension of energy is $M{{L}^{2}}{{T}^{-2}}$. The dimension of frequency is ${{T}^{-1}}$ and of angular momentum is $M{{L}^{2}}{{T}^{-1}}$. Dimensions of mass, distance and time are $M,L$and $T$respectively.
Formula used: Energy of a photon, $E=h\nu $, where h is the Planck's constant and $\nu $ is the frequency of the photon.
De Broglie Wavelength, $\lambda =\dfrac{h}{p}$, where $p$ is the momentum.
Complete step by step answer:
The energy of a photon of radiation is given by $E=h\nu $
Now, we can do the dimensional analysis on energy and frequency
$\left[ E \right]=\left[ h \right]\left[ \nu \right]$
So, we must substitute the values of dimensions of energy and frequency from the hint
$M{{L}^{2}}{{T}^{-2}}=\left[ h \right]{{T}^{-1}}$
$\begin{align}
& \Rightarrow \left[ h \right]=\dfrac{M{{L}^{2}}{{T}^{-2}}}{{{T}^{-1}}} \\
& \Rightarrow \left[ h \right]=M{{L}^{2}}{{T}^{-1}} \\
\end{align}$…(1)
Now, velocity is distance divided by time. This gives dimensions of velocity,
$v=L{{T}^{-1}}$
Also, momentum is mass multiplied by velocity,
This gives dimensions of momentum,
$\left[ p \right]=ML{{T}^{-1}}$
Furthermore, angular momentum has the same units as linear momentum multiplied with distance. So, we can find the dimensions of angular momentum.
$\left[ l \right]=M{{L}^{2}}{{T}^{-1}}$…(2)
Equating equations (1) and (2) we get,
$\left[ h \right]=\left[ l \right]$
Hence, the dimensions of Planck's constant are equal with the dimensions of angular momentum.
So, the correct answer is “Option D”.
Additional Information: The Planck's constant is divided by $2\pi $ to give a fundamental quantum mechanical constant $\hbar $.
The Planck's constant has value of $6.625\times {{10}^{-34}}$J-s in SI units.
The constant $\hbar $ appears in a number of important quantum mechanical equations like Schrödinger wave equation and the equation of temperature dependence of fermi level.
The Planck's constant was first defined by Max Plank to describe the phenomena of black body radiation.
It was further used by Bohr to describe the quantization of atomic orbitals by Neils Bohr.
It was also used by Einstein to describe the phenomena of photoelectric effect.
The expression $\hbar $ is used in the expression of the Heisenberg uncertainty principle and the quantized energy levels of harmonic oscillators.
It is right to assume that Planck's constant itself is used in much of the formulation of the entire quantum mechanics.
Note: Students can also solve this question by doing dimensional analysis on the equation of De Broglie wavelength shown in the formula used section. The question can also be solved by doing dimensional analysis on equation $E=h\dfrac{c}{\lambda }$, where c represents the velocity of light and $\lambda $ represents the wavelength of the wave.
Formula used: Energy of a photon, $E=h\nu $, where h is the Planck's constant and $\nu $ is the frequency of the photon.
De Broglie Wavelength, $\lambda =\dfrac{h}{p}$, where $p$ is the momentum.
Complete step by step answer:
The energy of a photon of radiation is given by $E=h\nu $
Now, we can do the dimensional analysis on energy and frequency
$\left[ E \right]=\left[ h \right]\left[ \nu \right]$
So, we must substitute the values of dimensions of energy and frequency from the hint
$M{{L}^{2}}{{T}^{-2}}=\left[ h \right]{{T}^{-1}}$
$\begin{align}
& \Rightarrow \left[ h \right]=\dfrac{M{{L}^{2}}{{T}^{-2}}}{{{T}^{-1}}} \\
& \Rightarrow \left[ h \right]=M{{L}^{2}}{{T}^{-1}} \\
\end{align}$…(1)
Now, velocity is distance divided by time. This gives dimensions of velocity,
$v=L{{T}^{-1}}$
Also, momentum is mass multiplied by velocity,
This gives dimensions of momentum,
$\left[ p \right]=ML{{T}^{-1}}$
Furthermore, angular momentum has the same units as linear momentum multiplied with distance. So, we can find the dimensions of angular momentum.
$\left[ l \right]=M{{L}^{2}}{{T}^{-1}}$…(2)
Equating equations (1) and (2) we get,
$\left[ h \right]=\left[ l \right]$
Hence, the dimensions of Planck's constant are equal with the dimensions of angular momentum.
So, the correct answer is “Option D”.
Additional Information: The Planck's constant is divided by $2\pi $ to give a fundamental quantum mechanical constant $\hbar $.
The Planck's constant has value of $6.625\times {{10}^{-34}}$J-s in SI units.
The constant $\hbar $ appears in a number of important quantum mechanical equations like Schrödinger wave equation and the equation of temperature dependence of fermi level.
The Planck's constant was first defined by Max Plank to describe the phenomena of black body radiation.
It was further used by Bohr to describe the quantization of atomic orbitals by Neils Bohr.
It was also used by Einstein to describe the phenomena of photoelectric effect.
The expression $\hbar $ is used in the expression of the Heisenberg uncertainty principle and the quantized energy levels of harmonic oscillators.
It is right to assume that Planck's constant itself is used in much of the formulation of the entire quantum mechanics.
Note: Students can also solve this question by doing dimensional analysis on the equation of De Broglie wavelength shown in the formula used section. The question can also be solved by doing dimensional analysis on equation $E=h\dfrac{c}{\lambda }$, where c represents the velocity of light and $\lambda $ represents the wavelength of the wave.
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